Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
        String[] line = bf.readLine().split(" ");
        int T = Integer.parseInt(line[0]);
        line = bf.readLine().split(" ");
        double[] in = new double[T];
        for (int i = 0; i < T; i++) {
            in[i] = Double.parseDouble(line[i]);
        }
        double[] dp = new double[T + 1];
        dp[0] = 1;
        for (int i = 1; i <= T; i++) {
            dp[i] = dp[i - 1] * in[i - 1];
            for (int j = i - 1; j > 0; j--) {
                dp[j] = dp[j] * (1 - in[i - 1]) + dp[j - 1] * in[i - 1];
            }
            dp[0] = dp[0] * (1 - in[i - 1]);
        }
        double ans = 0;
        for(int i = T/2+1; i<=T;i++) {
            ans += dp[i];
        }
        System.out.println(ans);
    }
}